FFTs in Graphics and Vision. The Fast Fourier Transform
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1 FFTs i Graphics ad Visio The Fast Fourier Trasform 1
2 Outlie The FFT Algorithm Applicatios i 1D Multi-Dimesioal FFTs More Applicatios Real FFTs 2
3 Computatioal Complexity To compute the movig dot-product of two periodic, -dimesioal arrays g[ ] ad h : 1. We eed to express g[ ] ad h[ ] i the basis v k [ ]: g = 1 k=0 g k v k ad h = 1 k=0 h k v k [ ] 2. We eed to multiply (ad scale) the coefficiets: g h = 1 k=0 g k h k v k [ ] 3. We eed to evaluate at every idex α: g h α = 1 k=0 g k h k v k [α] 3
4 Goal Give a -dimesioal array of complex values g[ ], we would like to compute the Fourier coefficiets of g[ ]: g = 1 k=0 g k v k [ ] where v k [ ] are the discrete samples of the complex expoetials at regularly spaced positios: v k = 1 eikθ 0,, e ikθ 1 4
5 Challege How ca we compute all Fourier coefficiets efficietly? 5
6 Challege How ca we compute all Fourier coefficiets? 6
7 Challege How ca we compute all Fourier coefficiets? Sice the vectors v k [ ] are orthoormal, we ca compute the k-th Fourier coefficiet of g[ ] by computig the dot-product: g k = g, v k = 1 j=0 1 = 1 j=0 g j v k j g j e ik2πj 7
8 Challege How ca we compute all Fourier coefficiets? Sice the vectors v k [ ] are orthoormal, we ca compute the k-th Fourier coefficiet of g[ ] by computig the dot-product: g k = g, v k 1 Computig oe coefficiet has complexity = g O() j v k j j=0 Computig all of them has = 1 1 complexity O( 2 ) g j e ik2πj j=0 8
9 Challege How ca we compute all Fourier coefficiets efficietly? 9
10 Approach (Divide ad Coquer) Key Idea: If we decompose the array ito the eve ad odd halves, we ca solve smaller problems ad the combie. 10
11 Approach (Divide ad Coquer) Key Idea: Cosider the 8-dimesioal array: g = (g 0, g 1, g 2, g 3, g 4, g 5, g 6, g 7 ) Ad cosider its eve/odd decompositio: g e = g 0, g 2, g 4, g 6 g o = (g 1, g 3, g 5, g 7 ) 11
12 Approach (Divide ad Coquer) Key Idea: g = g 0, g 1, g 2, g 3, g 4, g 5, g 6, g 7 g e = g 0, g 2, g 4, g 6 g o = (g 1, g 3, g 5, g 7 ) Now let s cosider the Fourier coefficiets: g k = iy 1 1 j=0 g j e ik2πj iy iy x x x e ik2π/8 e ik2π/4 e ik2π/4 12
13 Approach (Divide ad Coquer) Key Idea: g = g 0, g 1, g 2, g 3, g 4, g 5, g 6, g 7 g e = g 0, g 2, g 4, g 6 g o = (g 1, g 3, g 5, g 7 ) Now let s cosider the Fourier coefficiets: g k = iy 1 1 j=0 g j e ik2πj iy 0 th Order Coefficiets iy g[ ] g[2 ] g[2 +1] x x x g[0] g e [0] g o [0] 13
14 Approach (Divide ad Coquer) Key Idea: g = g 0, g 1, g 2, g 3, g 4, g 5, g 6, g 7 g e = g 0, g 2, g 4, g 6 g o = (g 1, g 3, g 5, g 7 ) Now let s cosider the Fourier coefficiets: g[5] g k = iy g[6] g[7] 1 1 j=0 g j e ik2πj iy g[6] 1 st Order Coefficiets iy g[7] g[4] g[0] x g[4] g[0] x g[5] g[1] x g[3] g[1] g[2] g[2] g[3] g[1] g e [1] g o [1] 14
15 Approach (Divide ad Coquer) Key Idea: g = g 0, g 1, g 2, g 3, g 4, g 5, g 6, g 7 g e = g 0, g 2, g 4, g 6 g o = (g 1, g 3, g 5, g 7 ) Now let s cosider the Fourier coefficiets: g k = iy g[3] g[7] 1 1 j=0 g j e ik2πj iy 2 d Order Coefficiets iy g[6] g[2] g[0] g[4] x g[6] g[2] g[0] g[4] x g[7] g[3] g[1] g[5] x g[5] g[1] g[2] g e [2] g o [2] 15
16 Approach (Divide ad Coquer) Key Idea: g = g 0, g 1, g 2, g 3, g 4, g 5, g 6, g 7 g e = g 0, g 2, g 4, g 6 g o = (g 1, g 3, g 5, g 7 ) Now let s cosider the Fourier coefficiets: g[7] g k = iy g[2] g[5] 1 1 j=0 g j e ik2πj iy g[2] 3 rd Order Coefficiets iy g[3] g[4] g[0] x g[4] g[0] x g[7] g[1] x g[1] g[3] g[6] g[6] g[4] g[3] g e [3] g o [3] 16
17 Approach (Divide ad Coquer) g[ ] g e [ ] g o [ ] For every frequecy: The eve compoets are i the same positio as the origial The odd oes are off by a fixed agle. g[5] g[4] g[ ] g[2 ] g[2 +1] g[0] x x x g[6] g[6] g[7] g[7] g[4] g[0] g[5] x x x g[3] g[1] g[2] g[2] g[3] g[3] g[7] g[1] g[6] g[2] g[0] g[4] g[6] g[2] g[0] g[4] g[7] g[3] g[1] g[5] x x x g[5] g[1] g[2] g[7] g[5] g[2] g[3] g[4] g[0] g[4] g[0] g[7] g[1] x x x g[1] g[3] g[6] g[6] g[4]
18 Approach (Divide ad Coquer) Assume that is eve ( = 2m) ad lets cosider the eve ad odd etries separately. That is, let g e [ ] ad g o [ ] be the m-dimesioal arrays: g e k = g 2k g o k = g[2k + 1] 18
19 Approach (Divide ad Coquer) g e k = g 2k g o k = g[2k + 1] The k-th Fourier coefficiet of g is: g k = 1 1 j=0 g j e ik2πj Splittig the summatio ito eve/odd terms: g k = m 1 1 j=0 2j ik2π g 2j e 2m 2j+1 ik2π + g 2j + 1 e 2m 19
20 Approach (Divide ad Coquer) g e k = g 2k g o k = g[2k + 1] Splittig the summatio ito eve/odd terms: g k = m 1 1 j=0 2j ik2π g 2j e 2m 2j+1 ik2π + g 2j + 1 e 2m Re-writig the expoet, we get: g k = m 1 1 j=0 g 2j e ik2πj m + g 2j + 1 e ik2πj m e ik2π 20
21 Approach (Divide ad Coquer) g e k = g 2k g o k = g[2k + 1] Re-writig the expoet, we get: g k = m 1 1 j=0 g 2j e ik2πj m + g 2j + 1 e ik2πj m e ik2π Pluggig i our eve/odd expressio gives: g k = m 1 1 j=0 g e j e ik2πj m + g o j e ik2πj m e ik2π 21
22 Approach (Divide ad Coquer) g e k = g 2k g o k = g[2k + 1] Pluggig i our eve/odd expressio gives: g k = m 1 1 j=0 g e j e ik2πj m + g o j e ik2πj m e ik2π Re-writig this equatio gives: g k = 1 m 1 j=0 g e j e ik2πj m + e ik2π 1 m 1 j=0 g o j e ik2πj m 22
23 Approach (Divide ad Coquer) g e k = g 2k g o k = g[2k + 1] Re-writig this equatio gives: g k = 1 m 1 j=0 g e j e ik2πj m + e ik2π 1 m 1 j=0 g o j e ik2πj m But this is a sum of Fourier coefficiets: g k = m g e k + g o k e ik2π 23
24 Approach (Divide ad Coquer) g e k = g 2k g o k = g[2k + 1] g k = m g e k + g o k e ik2π So, if we ca figure out the Fourier coefficiets of g e [ ] ad g o [ ], we ca combie them to get the Fourier coefficiets of g[ ]. Assumig that m is also eve, we ca repeat for g e [ ] ad g o [ ] to get their Fourier coefficiets. So if is a power of 2, we ca keep splittig, ot get a O( log ) algorithm. 24
25 Outlie The FFT Algorithm Applicatios i 1D Multi-Dimesioal FFTs More Applicatios Real FFTs 25
26 Applicatios of the FFT Movig Dot Products / Cross Correlatio g[ 2] g[ 1] g[0] g[1] g[2] h[ 2] h[ 1] h[0] h[1] h[2] 26
27 Applicatios of the FFT Movig Dot Products / Cross Correlatio g[ 2] g[ 1] g[0] g[1] g[2] h[ 2] h[ 1] h[0] h[1] h[2] 27
28 Applicatios of the FFT Movig Dot Products / Cross Correlatio g[ 2] g[ 1] g[0] g[1] g[2] h[ 2] h[ 1] h[0] h[1] h[2] 28
29 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio This is like cross-correlatio, except that flip the etries of the array h[ ] before cross-correlatig. g[ 2] h[2] g[ 2] h[ 2] g[ 1] h[1] g[ 1] h[ 1] g[0] h[0] g[0] h[0] g[1] h[ 1] g[1] h[1] g[2] h[ 2] g[2] h[2] 29
30 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio This is like cross-correlatio, except that flip the etries of the array h[ ] before cross-correlatig. g[ 2] h[2] g[ 2] h[ 2] g[ 1] h[1] g[ 1] h[ 1] g[0] h[0] g[0] h[0] g[1] h[ 1] g[1] h[1] g[2] h[ 2] g[2] h[2] 30
31 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio This is like cross-correlatio, except that flip the etries of the array h[ ] before cross-correlatig. g[ 2] h[2] g[ 2] h[ 2] g[ 1] h[1] g[ 1] h[ 1] g[0] h[0] g[0] h[0] g[1] h[ 1] g[1] h[1] g[2] h[ 2] g[2] h[2] 31
32 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio This is like cross-correlatio, except that flip the etries of the array h[ ] before cross-correlatig. Note: If h[ ] is symmetric (i.e. h k = h[k]) the the covolutio of g[ ] with h[ ] is equal to the crosscorrelatio of g[ ] with h[ ]. = g[ ] h[ ] g h = g h[ ] 32
33 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio 33
34 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio Give two polyomials: p x = a 0 + a 1 x + + a x q x = b 0 + b 1 x + + b x we ca represet the polyomials p(x) ad q(x) by (2 + 1)-dimesioal arrays: p x a,, a 1, a 0, a 1,, a q x b,, b 1, b 0, b 1,, b with: a = = a 1 = b = = b 1 = 0 34
35 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio The 0 th order coefficiet of the product is: a[ 2] a[ 1] a[0] a[1] a[2] b[ 2] b[ 1] b[0] b[1] b[2] 35
36 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio The 1 st order coefficiet of the product is: a[ 2] a[ 1] a[0] a[1] a[2] b[ 2] b[ 1] b[0] b[1] b[2] 36
37 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio The 2 d order coefficiet of the product is: a[ 2] a[ 1] a[0] a[1] a[2] b[ 2] b[ 1] b[0] b[1] b[2] 37
38 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio The coefficiets of the product ca be computed by covolvig the arrays correspodig to the coefficiets of the origial polyomials. a[ 2] a[ 1] a[0] a[1] a[2] b[ 2] b[ 1] b[0] b[1] b[2] 38
39 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio Big Iteger Multiplicatio Give a iteger, we ca treat it as a polyomial. Example:
40 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio Big Iteger Multiplicatio Give a iteger, we ca treat it as a polyomial. To multiply two itegers, we eed to figure out what the ew value i the 1s place, Example:
41 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio Big Iteger Multiplicatio Give a iteger, we ca treat it as a polyomial. To multiply two itegers, we eed to figure out what the ew value i the 1s place, the 10s place, Example:
42 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio Big Iteger Multiplicatio Give a iteger, we ca treat it as a polyomial. To multiply two itegers, we eed to figure out what the ew value i the 1s place, the 10s place, the 100s place, etc. will be. Example:
43 Applicatios of the FFT Movig Dot Products / Cross Correlatio Covolutio Polyomial Multiplicatio Big Iteger Multiplicatio Give a iteger, we ca treat it as a polyomial. To multiply two itegers, we eed to figure out what the ew value i the 1s place, the 10s place, the 100s place, etc. will be. So big iteger multiplicatio ca be implemeted as a covolutio. 43
44 Outlie The FFT Algorithm Applicatios i 1D Multi-Dimesioal FFTs More Applicatios Real FFTs 44
45 Multi-Dimesioal FFTs How do we compute the Fourier trasform of a multi-dimesioal sigal? Examples: Images Voxel Grids Etc. 45
46 2D FFTs For regularly sampled, grids, the irreducible represetatios are spaed by the orthogoal basis {v lm } where: v lm j k = 1 2 eil2πj e im2πk To see this, cosider the shift by (α, β): ρ α,β v lm [ ] j k = v lm j α k β = 1 j α eil2π 2 im2π k β e = e il2πα e im2πβ v lm [j][k] 46
47 2D FFTs How ca we compute the 2D Fourier coefficiets efficietly? To do this, we will leverage the fact that the 2D basis vectors ca be expressed as the product of 1D basis vectors. Settig v l [ ] to be the -dimesioal array: v l j = 1 eil2πj we get: v lm j k = v l j v m [k] 47
48 2D FFTs To compute the l, m -th Fourier coefficiet of a ( )-dimesioal grid g [ ], we compute the dot-product of g [ ] with v lm [ ]: g l m = g, v lm = 1 1 j=0 g j k v lm j k k=0 1 1 = 1 2 j=0 1 = 1 j=0 k=0 1 g j k e im2πk 1 k=0 g j k e im2πk e il2πj e il2πj 48
49 2D FFTs g l [m] = 1 1 j=0 1 1 k=0 g j k e im2πk e il2πj The iterior sum is a 1D Fourier trasform! I particular, settig g j [ ] to be the 1D array: g j k = g j k we get: g l m = 1 1 j=0 1 1 k=0 g j k e im2πk e il2πj 1 = 1 j=0 g j m e il2πj 49
50 2D FFTs g l [m] = 1 1 j=0 g j m e il2πj But this sum is also a 1D Fourier trasform! Thus, to compute the Fourier coefficiets, we: 1. Compute the Fourier coefficiets of each row 2. Compute the Fourier coefficiets of each colum Ad the total complexity of this operatio is: O( 2 log ) 50
51 Outlie The FFT Algorithm Applicatios i 1D Multi-Dimesioal FFTs More Applicatios Gaussia Smoothig Up-Samplig Differetiatio Boudary Detectio Gaussia Sharpeig Real FFTs 51
52 Gaussia Smoothig Give a grid of values, we would like to smooth the grid. 52
53 Gaussia Smoothig Give a grid of values, we would like to smooth the grid. To do this we eed: g [ ] : The iitial grid h [ ] : The smoothig filter, usually a Gaussia: g j k = e λ j2 +k 2 j,k=0 1 e λ j2 +k 2 with /2 < j, k /2. g h[ ] : The Gaussia-smoothed grid 53
54 Up-Samplig Give a -dimesioal array, we would like to extrapolate the array to a 2-dimesioal array. Oe way to do this is to fit a cotiuous fuctio to the origial array ad the sample at 2 regular samples. 54
55 Up-Samplig How do we geerate a cotiuous fuctio from a set of samples? Recall that the Fourier decompositio of g[ ] is: g j = 1 1 k=0 g k e ik2πj 55
56 Up-Samplig How do we geerate a cotiuous fuctio from a set of samples? We ca fit a cotiuous fuctio to the data by replacig the discrete idex 0 j with a cotiuous idex 0 s : g j = 1 1 k=0 1 g k e ik2πj g(s) = 1 k=0 g k e ik2πs Sice at iteger values of j we have: g j = g j the cotiuous fuctio will iterpolate the discrete samples. 56
57 Up-Samplig Word of Warig: Recall that for iteger values of j, the complex expoetial satisfies the coditio: e ik2πj = e i k+ 2πj Thus, if we were to fit a fuctio: g j = 1 1 k=0 1 g k e ik2πj g s = 1 k=0 g k e i(k+)2πs we would also get a cotiuous fuctio that iterpolates the discrete samples. 57
58 Up-Samplig Word of Warig: The differece is i how the array is iterpolated: g s = 1 1 k=0 g k e ik2πs g s = 1 1 g k e i(k+)2πs k=0 58
59 Up-Samplig Word of Warig: Extrapolatig the discrete samples is a udercostraied problem, ad so there are may differet solutios. I practice, we would like the smoothest possible fit, so we would like to miimize the cotributio of high frequecy terms g s = 1 1 k=0 g k e ik2πs g s = 1 1 g k e i(k+)2πs k=0 59
60 Up-Samplig Word of Warig: Thus, the best fit is obtaied usig the fuctio: g s = /2 1 k= /2+1 g k e ik2πs A simple algorithm for implemetig this is: 1. Compute the Fourier coefficiets of g[ ] 2. Geerate a array of 2 Fourier coefficiets: h k = g[k] /2 k /2 0 otherwise 3. Compute the iverse Fourier trasform to get back the 2-dimesioal array h[ ] 60
61 Differetiatio Give a -dimesioal array g[ ], how do we compute the derivative of g[ ]? Fiite Differeces: Defie the derivative at some idex j as the average of the discrete left ad right derivatives: g g j + 1 g j + g j g j 1 j = 2 g j + 1 g j 1 = 2 61
62 Differetiatio Give a -dimesioal array g[ ], how do we compute the derivative of g[ ]? Fiite Differeces: Defie the derivative at some idex j as the average of the discrete left ad right derivatives: g g j + 1 g j 1 j = 2 Cotiuous Differetiatio: Fit a cotiuous fuctio to the samples ad take the derivative of the cotiuous fuctio. 62
63 Differetiatio Cotiuous Differetiatio: Fit a cotiuous fuctio to the samples ad take the derivative of the cotiuous fuctio. If we fit a cotiuous fuctio to g[ ] by: g s = /2 1 k= /2+1 g k e ik2πs The usig the fact that s eλs = λe λs we get: g s = /2 1 k= /2+1 g k ik2π e ik2πs 63
64 Differetiatio Cotiuous Differetiatio: g s = /2 1 k= /2+1 g k ik2π e ik2πs Thus, we ca obtai the values of the cotiuous derivative by multiplyig the k-th Fourier coefficiet of g[ ] by (ik2π/): g k = ik2π g[k] 64
65 Boudary Detectio To compute the boudary of a grid g [ ], we would like to measure how much the grid g [ ] is chagig at every idex. Specifically, we would like to defie a grid h [ ] such that h j [k] measures the extet of chage of g [ ] at the idex (j, k). g [ ] h [ ] 65
66 Boudary Detectio Gradiet Method: Oe way to measure the extet chage is by computig the gradiet of g [ ] at every poit ad settig: h j k = g j k To compute the gradiet, we eed to compute the partial derivatives of g [ ] at every poit. 66
67 Boudary Detectio Gradiet Method: This ca be doe with the FFT: 1. Compute the Fourier coefficiets of g [ ] 2. Geerate the two grids correspodig to the Fourier coefficiets of the partial derivatives: g x j k = g j k ij2π/ g y j k = g j k (ik2π/) 3. Compute the iverse Fourier trasforms to get the grids of partial derivatives g x [ ] ad g y [ ]. 4. Set h [ ] to be the grid of gradiet legths: h j k = g x j k 2 + g y j k 2 67
68 Boudary Detectio Laplacia Method: A alterate way to measure the rate of chage is to compute the differece betwee the origial grid ad a smoothed versio of the grid. 68
69 Boudary Detectio Laplacia Method: A alterate way to measure the rate of chage is to compute the differece betwee the origial grid ad a smoothed versio of the grid. A measure of this differece ca be obtaied by computig the Laplacia (the sum of umixed) partial derivatives: Δf = f xx + f yy 69
70 Boudary Detectio Laplacia Method: This ca be doe with the FFT: 1. Compute the Fourier coefficiets of g [ ] 2. Geerate the grid correspodig to the Fourier coefficiets of the Laplacia: g xx + g yy j k = g j k ij2π/ 2 + ik2π/ 2 = g j k j 2 + k 2 4π 2 / 2 3. Set h [ ] to be the iverse Fourier trasform of the Laplacia Fourier coefficiets. 70
71 Gaussia Sharpeig How do we udo the effects of Gaussiasmoothig a grid? We perform Gaussia smoothig of g [ ] by: 1. Computig the Fourier trasforms of g [ ] ad the Gaussia filter h [ ] 2. Multiplyig the Fourier coefficiets of g [ ] by the Fourier coefficiets of h [ ] 3. Computig the iverse Fourier trasform 71
72 Gaussia Sharpeig How do we udo the effects of Gaussiasmoothig a grid? We u-perform Gaussia smoothig of g [ ] by: 1. Computig the Fourier trasforms of g [ ] ad the Gaussia grid h [ ] 2. Multiplyig the Fourier coefficiets of g [ ] by the reciprocals of the Fourier coefficiets of h [ ] 3. Computig the iverse Fourier trasform This is doable as log as the Fourier coefficiets of h [ ] are o-zero. This is umerically stable as log as the Fourier coefficiets of h [ ] are ot too small. 72
73 Outlie The FFT Algorithm Applicatios i 1D Multi-Dimesioal FFTs More Applicatios Real FFTs 73
74 Real FFTs So far, we have cosidered the Fourier trasform of complex valued fuctios. What happes whe the values of the fuctio are all real? g θ g θ = 0 74
75 Real FFTs If we write out the fuctio g i terms of its Fourier decompositio, we get: g θ = 1 2π k= g k e ikθ Usig the fact that g g = 0 we get: 0 = 1 2π k= g k e ikθ But this ca be re-writte as: 0 = 1 2π k= g k e ikθ k= k= g k e ikθ g k e ikθ 75
76 Real FFTs 0 = 1 2π k= g k e ikθ k= Simplifyig this equatio we get: 0 = k= g k g k g k e ikθ e ikθ But sice complex expoetials are liear liearly idepedet, this implies that all the Fourier coefficiets are equal to zero: 0 = g k g k g k = g k 76
77 Real FFTs g k = g k Thus whe the fuctio g is real, the k-th ad ( k)-th Fourier coefficiet are cojugate. 77
78 Real FFTs Although this discussio holds true for real fuctios, a similar argumet shows that for realvalued, -dimesioal arrays g[ ], the Fourier coefficiets have the property that: g k = g k 78
79 Real FFTs For real-valued arrays: I 1D we have: g j = g j I 2D we have: g j [k] = g j [ k] I 3D we have: g j k [l] = g j k [ l] So whe the array is real-valued, we oly have to compute ad store half of the coefficiets. 79
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